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Theorem 2.26. (Implicit function theorem) Let \( f : U \times V \rightarrow {\mathbb{R}}^{m} \) be a \( {C}^{1} \) mapping, where \( U \subseteq {\mathbb{R}}^{n} \) and \( V \subseteq {\mathbb{R}}^{m} \) are open sets. Let \( \left( {{x}_{0},{y}_{0}}\right) \in U \times V \) be a point such that \( f\left( {{x}_{0},{y}... | Proof. Assume without loss of generality that \( {x}_{0} = 0 \) and \( {y}_{0} = 0 \), by considering the function \( \left( {x, y}\right) \mapsto f\left( {x + {x}_{0}, y + {y}_{0}}\right) - f\left( {{x}_{0},{y}_{0}}\right) \) if necessary. Let \( f\left( x\right) = \left( {{f}_{1}\left( {x, y}\right) ,\ldots ,{f}_{m}\... | Yes |
Corollary 2.27. (Inverse function theorem) Let \( f \) be a \( {C}^{1} \) map from a neighborhood of \( {x}_{0} \in {\mathbb{R}}^{n} \) into \( {\mathbb{R}}^{n} \). If \( {Df}\left( {x}_{0}\right) \) is nonsingular, then there exist neighborhoods \( U \ni {x}_{0} \) and \( V \ni \) \( {y}_{0} = f\left( {x}_{0}\right) \... | Proof. Define the function \( F\left( {x, y}\right) = f\left( x\right) - y \), and note that \( {D}_{x}F\left( {{x}_{0}, y}\right) = \) \( {Df}\left( {x}_{0}\right) \) is nonsingular. Apply Theorem 2.26 to \( F \) . | No |
Theorem 2.29. (Lyusternik) Let \( f : U \rightarrow {\mathbb{R}}^{m} \) be a \( {C}^{1} \) map, where \( U \subset {\mathbb{R}}^{n} \) is an open set. Let \( M = {f}^{-1}\left( {f\left( {x}_{0}\right) }\right) \) be the level set of a point \( {x}_{0} \in U \). If the derivative \( {Df}\left( {x}_{0}\right) \) is a lin... | Proof. We may assume that \( {x}_{0} = 0 \) and \( f\left( {x}_{0}\right) = 0 \), by considering the function \( x \mapsto f\left( {x + {x}_{0}}\right) - f\left( {x}_{0}\right) \) if necessary. Define \( A \mathrel{\text{:=}} {Df}\left( 0\right) \). The proof of the inclusion \( {T}_{M}\left( 0\right) \subseteq \operat... | Yes |
Lemma 2.31. Let \( {S}^{n} \) be the space of \( n \times n \) symmetric matrices, \( A \in {S}^{n} \) nonsingular, and let \( {S}_{A}^{n} \) be the vector space of \( n \times n \) matrices \( X \) such that \( {AX} \) is symmetric. The quadratic map \[ {q}_{A} : {S}_{A}^{n} \rightarrow {S}^{n}\;\text{ defined by }{q}... | Proof. We have \[ q\left( {I + {tH}}\right) \mathrel{\text{:=}} {q}_{A}\left( {I + {tH}}\right) = \left( {I + t{H}^{T}}\right) A\left( {I + {tH}}\right) \] \[ = A + t\left( {{H}^{T}A + {AH}}\right) + {t}^{2}{H}^{T}{AH} = A + {2tAH} + {t}^{2}A{H}^{2}, \] so that \( {Dq}\left( I\right) \left( H\right) = {2AH} \) . The ma... | Yes |
Theorem 2.32. (Morse’s lemma) Let \( k \geq 1 \) and \( f : U \rightarrow \mathbb{R} \) be a \( {C}^{2 + k} \) function on an open set \( U \subseteq {\mathbb{R}}^{n} \) . If \( {x}_{0} \in U \) is a nondegenerate critical point of \( f \), then there exist open neighborhoods \( V \ni {x}_{0} \) and \( W \ni 0 \) in \(... | Proof. We may assume without any loss of generality that \( U \) is a convex set, \( {x}_{0} = 0 \), and \( f\left( 0\right) = 0 \) . Let \( 0 \neq x \in U \), and define \( \alpha \left( t\right) \mathrel{\text{:=}} f\left( {tx}\right) \) . We have \[ \alpha \left( 1\right) = \alpha \left( 0\right) + {\alpha }^{\prime... | Yes |
Corollary 2.33. Let \( f : U \rightarrow \mathbb{R} \) be a \( {C}^{2 + k} \) function as in Theorem 2.32, and let \( {x}_{0} \in U \) be a nondegenerate critical point of \( f \) such that the Hessian matrix \( A = {Df}\left( {x}_{0}\right) \) has \( k\left( {0 \leq k \leq n}\right) \) positive and \( n - k \) negativ... | Proof. Let \( A \mathrel{\text{:=}} {Df}\left( {x}_{0}\right) \) have the spectral decomposition \( A = {U}^{T}{\Lambda U} \), where \( \Lambda = \operatorname{diag}\left( {{\lambda }_{1},\ldots ,{\lambda }_{k},\ldots ,{\lambda }_{n}}\right) \) with \( {\lambda }_{i} > 0 \) for \( i \leq k \) and \( {\lambda }_{i} < 0 ... | Yes |
Theorem 2.39. Let \( f : E \rightarrow \mathbb{R} \cup \{ \pm \infty \} \) . The following are equivalent:\n\n(a) \( f \) is lower semicontinuous (upper semicontinuous) on \( E \) ,\n\n(b) \( \operatorname{epi}\left( f\right) \) (hypo \( \left( f\right) \) ) is a closed subset of \( E \times \mathbb{R} \) ,\n\n(c) The ... | Proof. We prove the theorem only for a lower semicontinuous function, since the upper semicontinuous case follows immediately.\n\n(a) implies (b): Let \( \left( {{x}_{n},{y}_{n}}\right) \) be a sequence in \( \operatorname{epi}\left( f\right) \) converging to a point \( \left( {x, y}\right) \) . Since \( f \) is lower ... | Yes |
Corollary 2.40. If the functions \( f, g : E \rightarrow \mathbb{R} \cup \{ + \infty \} \) are lower semicontinuous, then so is \( f + g \) . | Proof. We claim that\n\n\[ \n\{ x : f\left( x\right) + g\left( x\right) > t\} = { \cup }_{\alpha \in \mathbb{R}}\left( {\{ x : f\left( x\right) > t - \alpha \} \cap \{ x : g\left( x\right) > \alpha \} }\right) .\n\]\n\nIf \( f\left( x\right) + g\left( x\right) = t + {2\varepsilon } > t \) and \( g\left( x\right) = \alp... | Yes |
Theorem 2.41. Let \( f : E \rightarrow \mathbb{R} \cup \{ \infty \} \) be a lower semicontinuous function defined on a metric space \( E \) . If \( f \) has a nonempty compact sublevel set, \[ {l}_{\alpha }\left( f\right) \mathrel{\text{:=}} \{ x \in E : f\left( x\right) \leq \alpha \} \] then \( f \) achieves its glob... | Proof. Let \( \left\{ {x}_{n}\right\} \) be a minimizing sequence for \( f \), that is, \[ f\left( {x}_{n}\right) \searrow \inf \{ f\left( x\right) : x \in E\} = : \mathop{\inf }\limits_{E}f. \] Clearly, there exists an integer \( N \) such that \( {x}_{n} \in {l}_{\alpha }\left( f\right) \) for all \( n \geq N \) . Si... | Yes |
Corollary 2.43. Let \( f : D \rightarrow \mathbb{R} \) be a lower semicontinuous function defined on a topological space \( D \) .\n\n(a) If \( D \) is compact, or\n\n(b) \( D \) is a subset of a finite-dimensional normed vector space \( E \) and \( f \) is coercive,\n\nthen \( f \) achieves a global minimum on \( D \)... | Proof. In either case, all sublevel sets of \( f \) are compact. In (ii), this follows from the fact that the sublevel sets of \( f \) are closed and bounded, hence compact. | No |
Theorem 3.2. (Ekeland’s \( \epsilon \) -variational principle) Let \( \left( {M, d}\right) \) be a complete metric space, and let \( f : M \rightarrow \mathbb{R} \cup \{ + \infty \} \) be a proper lower semicontinuous function that is bounded from below. Then for every \( \epsilon > 0,\lambda > 0 \), and \( x \in M \) ... | Proof. It suffices to prove the theorem for \( \lambda = 1 \) and \( \epsilon = 1 \) ; the general case follows by replacing the distance function \( d \) by the equivalent distance function \( d/\lambda \) and the function \( f \) by the function \( f/\epsilon \) . It follows from Theorem 3.1 that there exists a \( d ... | Yes |
Corollary 3.5. Let \( f : X \rightarrow \mathbb{R} \) be a function on a Banach space \( X \) that is Gâteaux differentiable, lower semicontinuous, and bounded from below. Let \( \epsilon > 0 \), and let \( x \in X \) be a point such that\n\n\[ f\left( x\right) \leq \mathop{\inf }\limits_{X}f + \epsilon . \]\n\nThen th... | Proof. Theorem 3.2 gives a point \( {x}_{\epsilon } \) satisfying the first two conditions above. To prove the third condition, note that for an arbitrary direction \( d \in X \) , \( \parallel d\parallel = 1 \), we have\n\n\[ t\left\langle {\nabla f\left( {x}_{\epsilon }\right), d}\right\rangle + o\left( t\right) = f\... | Yes |
Theorem 3.6. Let \( f : X \rightarrow \mathbb{R} \) be a function on a Banach space \( X \) that is Gâteaux differentiable, lower semicontinuous, bounded from below, but that is not coercive. Define\n\n\[ \alpha \mathrel{\text{:=}} \mathop{\lim }\limits_{{\parallel x\parallel \rightarrow \infty }}f\left( x\right) = \ma... | Proof. Let \( \left( {{M}_{n}, d}\right) \) be the complete metric space such that \( {M}_{n} \mathrel{\text{:=}} \{ x \in X \) : \( \parallel x\parallel \geq n\} \) and the distance function \( d \) is given by the norm on \( X \) . Let \( \begin{Vmatrix}{y}_{n}\end{Vmatrix} \geq \) \( n + 2 \) be a point satisfying\n... | Yes |
Theorem 3.8. (Banach fixed point theorem) A contractive mapping \( \varphi \) : \( M \rightarrow M \) on a complete metric space \( \left( {M, d}\right) \) has a unique fixed point. | Proof. Define the function \( f\left( x\right) \mathrel{\text{:=}} d\left( {x,\varphi \left( x\right) }\right) \), and choose \( \epsilon \in \left( {0,1 - \alpha }\right) \) . Theorem 3.2 implies that there exists a point \( \bar{x} \in M \) such that\n\n\[ f\left( \bar{x}\right) \leq f\left( x\right) + {\epsilon d}\l... | Yes |
Theorem 3.10. (Borwein-Preiss Variational Principle) Let \( \\left( {M, d}\\right) \) be a complete metric space and let \( f : M \\rightarrow \\mathbb{R} \\cup \\{ + \\infty \\} \) be a proper lower semicontinuous function, bounded from below. Let \( \\rho \) be a gauge-type function and \( {\\left\\{ {\\delta }_{k}\\... | The proof of this theorem resembles that of Theorem 3.2, which the interested reader can find in the book [43]. We will instead give a version of this theorem in finite-dimensional Euclidean spaces that is much simpler to prove, but which is sufficient for our purposes. | No |
Theorem 3.11. (Smooth variational principle in finite-dimensional spaces) Let \( f : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \cup \{ + \infty \} \) be a proper lower semicontinuous function, bounded from below. Let \( \lambda \) and \( p \geq 1 \) .\n\nIf \( \epsilon > 0 \) and \( x \in {\mathbb{R}}^{n} \) satisfies\n\... | Proof. Note that the function \( g\left( z\right) \mathrel{\text{:=}} f\left( z\right) + \frac{\epsilon }{{\lambda }^{p}}\parallel z - x{\parallel }^{p} \) is coercive, and so has a minimizer \( {x}_{\epsilon } \in {\mathbb{R}}^{n} \) . This proves the second assertion of the theorem. If we set \( z = x \) in this ineq... | Yes |
Corollary 3.12. If a \( {C}^{2} \) function \( f : {\mathbb{R}}^{k} \rightarrow \mathbb{R} \) is bounded from below, then there exists a sequence \( {\left\{ {x}_{n}\right\} }_{1}^{\infty } \) in \( {\mathbb{R}}^{k} \) satisfying the properties\n\n\[ f\left( {x}_{n}\right) \rightarrow \mathop{\inf }\limits_{{\mathbb{R}... | Proof. Let \( p = 2 \) and \( \lambda = 1 \) in Theorem 3.11. Let \( x,\epsilon > 0 \), and \( {x}_{\epsilon } > 0 \) be as in that theorem. Note that for \( d \in {\mathbb{R}}^{k} \), the point \( {x}_{\epsilon } \) satisfies \( \begin{Vmatrix}{{x}_{\epsilon } - x}\end{Vmatrix} \leq 1 \) and the conditions\n\n\[ f\lef... | Yes |
Theorem 3.14. Define the function\n\n\[ f\left( x\right) = \ln \mathop{\sum }\limits_{{i = 1}}^{m}{e}^{\left\langle {a}_{i}, x\right\rangle } \]\n\nwhere \( {\left\{ {a}_{i}\right\} }_{1}^{m} \) are vectors in \( {\mathbb{R}}^{n} \). The following statements are equivalent:\n\n(a) \( f\left( x\right) \) is bounded from... | Proof. The proofs of \( \left( b\right) \Rightarrow \left( c\right) \) and \( \left( c\right) \Rightarrow \left( a\right) \) are trivial.\n\nTo prove \( \left( a\right) \Rightarrow \left( b\right) \), note that Corollary 3.5 gives a sequence \( \left\{ {x}_{n}\right\} \) satisfying\n\n\[ \nabla f\left( {x}_{n}\right) =... | No |
Corollary 3.16. (Farkas’s lemma, homogeneous version) Let \( {\left\{ {a}_{i}\right\} }_{1}^{m} \) and \( c \) be vectors in \( {\mathbb{R}}^{n} \) . The following statements are equivalent:\n\n(a) if \( x \) satisfies \( \left\langle {{a}_{i}, x}\right\rangle \leq 0, i = 1,\ldots, m \), then it also satisfies \( \lang... | Proof. The corollary follows immediately from Theorem 3.15 by noting that the validity of (a) is equivalent to the inconsistency of the linear inequality system\n\n\[ \langle - c, x\rangle < 0,\left\langle {{a}_{i}, x}\right\rangle \leq 0, i = 1,\ldots, m. \] | Yes |
Theorem 3.17. (Motzkin's transposition theorem, affine version) Let \( {\left\{ {a}_{i}\right\} }_{1}^{l},{\left\{ {b}_{j}\right\} }_{1}^{m},{\left\{ {c}_{k}\right\} }_{1}^{p} \) be vectors in \( {\mathbb{R}}^{n} \), and let \( {\left\{ {\alpha }_{i}\right\} }_{1}^{l},{\left\{ {\beta }_{j}\right\} }_{1}^{m},{\left\{ {\... | Proof. Note that the inconsistency of the system (3.7) is equivalent to that of the homogenized system \( \left\langle {{a}_{i}, x}\right\rangle < t{\alpha }_{i},\left\langle {{b}_{j}, x}\right\rangle \leq t{\beta }_{j},\left\langle {{c}_{k}, x}\right\rangle = t{\gamma }_{k}, t > 0 \), that is, of the system\n\n\[ \lan... | Yes |
Corollary 3.18. (Farkas’s lemma, affine version) Let\n\n\\[ \n\\left\\langle {{a}_{i}, x}\\right\\rangle \\leq {\\alpha }_{i}, i = 1,\\ldots, m\n\\]\n\n(3.10)\n\nbe a consistent system of linear inequalities, where \\( {\\left\\{ {a}_{i}\\right\\} }_{1}^{m} \\subset {\\mathbb{R}}^{n} \\) . The following statements are ... | Proof. Note that the validity of (a) is equivalent to the inconsistency of the linear inequality system\n\n\\[ \n\\langle - c, x\\rangle < - \\gamma ,\\left\\langle {{a}_{i}, x}\\right\\rangle \\leq {\\alpha }_{i}, i = 1,\\ldots, m.\n\\]\n\nIt follows from Theorem 3.17 that there exist nonnegative multipliers \\( 0 \\n... | Yes |
Lemma 3.20. Let \( T : X \rightarrow Y \) be a continuous linear mapping from a Banach space \( X \) onto a Banach space \( Y \) . Then there exists a constant \( \tau > 0 \) such that \( \tau {\bar{B}}_{Y} \subseteq \overline{A\left( {\bar{B}}_{X}\right) } \), where \( {\bar{B}}_{X} = \{ x \in X : \parallel x\parallel... | Proof. Since \( A \) is onto,\n\n\[ Y = A\left( X\right) = A\left( {{ \cup }_{n = 1}^{\infty }n{\bar{B}}_{X}}\right) = { \cup }_{n = 1}^{\infty }A\left( {n{\bar{B}}_{X}}\right) = { \cup }_{n = 1}^{\infty }{nA}\left( {\bar{B}}_{X}\right) .\n\]\n\nIt follows from the Baire category theorem that at least one set \( \overl... | Yes |
Theorem 3.21. (Graves’s theorem) Let \( X \) and \( Y \) be Banach spaces, \( r > 0 \), and let \( f : r{\bar{B}}_{X} \rightarrow Y \) be a mapping such that \( f\left( 0\right) = 0 \). Let \( A : X \rightarrow Y \) be a continuous linear mapping onto \( Y \) satisfying \( \tau {\bar{B}}_{Y} \subseteq \overline{A\left(... | Proof. Define \( D = r{\bar{B}}_{X} \), and for each point \( y \in Y \), define the function \( {f}_{y} \) on \( D \) ,\n\n\[ {f}_{y}\left( x\right) \mathrel{\text{:=}} \parallel f\left( x\right) - y\parallel . \]\n\nWe first claim that\n\n\[ \left| {\nabla {f}_{y}}\right| \left( x\right) \geq c > 0\text{ for all }x \... | Yes |
Corollary 3.22. (Open mapping theorem) Let \( X \) and \( Y \) be Banach spaces and let \( A : X \rightarrow Y \) be a continuous linear mapping onto \( Y \) . Then \( A \) is an open mapping, that is, if \( O \subseteq X \) is open, then \( A\left( O\right) \) is open in \( Y \) . | Proof. Applying Theorem 3.21 with \( f = A, y = 0 \), and \( \delta = 0 \), we obtain\n\n\[ \n{\tau d}\left( {x,{A}^{-1}0}\right) \leq \parallel {Ax}\parallel \n\]\n\nfor all \( x \) with small enough norm, and hence for all \( x \), by the homogeneity of the above inequality. If \( y = {Ax} \) satisfies \( \parallel y... | Yes |
Theorem 3.23. (Lyusternik’s theorem) Let \( X \) and \( Y \) be Banach spaces, \( U \subseteq X \) an open set, and \( f : U \rightarrow Y \) . Let \( {T}_{M}\left( {x}_{0}\right) \) be the tangent cone of the level set \( M = {f}^{-1}\left( {f\left( {x}_{0}\right) }\right) \) at the point \( {x}_{0} \in U \) .\n\nIf \... | Proof. As in the proof of Theorem 2.29, we may assume that \( {x}_{0} = 0 \) and \( f\left( {x}_{0}\right) = 0 \) . Define \( A = {Df}\left( 0\right) \) . The proof of the inclusion \( {T}_{M}\left( 0\right) \subseteq \operatorname{Ker}A \) is the standard one given there, namely if \( d \in {T}_{M}\left( 0\right) \), ... | Yes |
Theorem 3.26. (Implicit function theorem) Let \( X, Y \) be Banach spaces, \( {x}_{0} \in X,{y}_{0} \in Y \), and \( f \) a \( {C}^{1} \) map in a neighborhood of \( \left( {{x}_{0},{y}_{0}}\right) \) . Define \( {w}_{0} = f\left( {{x}_{0},{y}_{0}}\right) \n\nIf \( {D}_{y}f\left( {{x}_{0},{y}_{0}}\right) : Y \rightarro... | Proof. Define the map \( F\left( {x, y}\right) = \left( {x, f\left( {x, y}\right) }\right) \) . At a point \( {z}_{0} = \left( {{x}_{0},{y}_{0}}\right) \), it follows from Taylor's formula\n\n\[ \nF\left( {{x}_{0} + {th},{y}_{0} + {tk}}\right) = \left( {{x}_{0} + {th}, f\left( {{x}_{0} + {th},{y}_{0} + {tk}}\right) }\r... | Yes |
Lemma 4.2. Let \( A \) be a nonempty set in \( E \) . Then \( \operatorname{aff}\left( A\right) \) is an affine set, in fact, the smallest affine set containing \( A \) . | Proof. It is clear that \( A \subseteq \operatorname{aff}\left( A\right) \) . Let us show that \( \operatorname{aff}\left( A\right) \) is an affine set. If \( t \in \mathbb{R} \) and \( u, v \in \operatorname{aff}\left( A\right) \) have the forms\n\n\[ u = \mathop{\sum }\limits_{{i = 1}}^{k}{\lambda }_{i}{x}_{i}, v = \... | Yes |
Lemma 4.3. If \( A \subseteq E \) is an affine set, then \( \operatorname{aff}\left( A\right) = A \), that is, all affine combinations of elements from \( A \) lie in \( A \) : | \[ {\left\{ {\lambda }_{i}\right\} }_{1}^{k} \subset \mathbb{R},\;\mathop{\sum }\limits_{{i = 1}}^{k}{\lambda }_{i} = 1,\;{\left\{ {x}_{i}\right\} }_{1}^{k} \subset A\; \Rightarrow \;\mathop{\sum }\limits_{{i = 1}}^{k}{\lambda }_{i}{x}_{i} \in A. \] | Yes |
Theorem 4.4. If \( A \subseteq E \) is an affine subset of \( E \), and \( a \in A \) is an arbitrary point, then\n\n\[ L \mathrel{\text{:=}} A - a = \{ y - a : y \in A\} \]\n\n is a linear subspace of \( E \), which is independent of \( a \in A \) ; consequently,\n\n\[ A = a + L,\;\text{ and }\;L = A - A = \{ y - z : ... | Proof. Let us first prove that \( L \) is a linear subspace of \( E \) . Let \( \left\{ {{x}_{1},{x}_{2}}\right\} \subseteq L \) and \( \left\{ {{\alpha }_{1},{\alpha }_{2}}\right\} \subset \mathbb{R} \) . Writing \( {x}_{i} = {y}_{i} - a,{y}_{i} \in A \), we have\n\n\[ y \mathrel{\text{:=}} a + {\alpha }_{1}{x}_{1} + ... | Yes |
Lemma 4.6. Let \( A \subseteq E \) be an affine subset of \( E \) with the corresponding linear subspace \( L = A - A \) . A subset \( {\left\{ {x}_{i}\right\} }_{1}^{l} \) of \( A \) is affinely independent if and only if \[ \mathop{\sum }\limits_{1}^{l}{\alpha }_{i}{x}_{i} = 0,\;\mathop{\sum }\limits_{1}^{l}{\alpha }... | Proof. Notice that \( x = \mathop{\sum }\limits_{1}^{l}{\alpha }_{i}{x}_{i} \) and \( x = \mathop{\sum }\limits_{1}^{l}{\beta }_{i}{x}_{i} \) are two distinct affine combinations if and only if \( 0 = \mathop{\sum }\limits_{1}^{l}\left( {{\beta }_{i} - {\alpha }_{i}}\right) {x}_{i} \) is a nontrivial linear combination... | Yes |
Theorem 4.7. Let \( F : A \rightarrow B \) be an affine map between affine sets \( A \subseteq {E}_{1} \) and \( B \subseteq {E}_{2} \) . Then \( F \) preserves affine combinations, that is,\n\n\[ F\left( {\mathop{\sum }\limits_{{i = 1}}^{k}{\alpha }_{i}{x}_{i}}\right) = \mathop{\sum }\limits_{{i = 1}}^{k}{\alpha }_{i}... | Proof. The graph \( C = \{ \left( {x, y}\right) : x \in A, y = F\left( x\right) \} \) is an affine set in \( {E}_{1} \times {E}_{2} \) . It follows from Lemma 4.3 that if \( \mathop{\sum }\limits_{1}^{k}{\alpha }_{i} = 1 \), then \( \mathop{\sum }\limits_{1}^{k}{\alpha }_{i}\left( {{x}_{i}, F\left( {x}_{i}\right) }\rig... | Yes |
Intersections of convex sets are convex: if \( {\left\{ {C}_{\gamma }\right\} }_{\gamma \in \Gamma } \) is a family of convex sets in \( E \), then \( { \cap }_{\gamma \in \Gamma }{C}_{\gamma } \) is a convex set. | Proof. These statements are all easy to prove; we prove only (a). Let \( x, y \in \) \( C \mathrel{\text{:=}} { \cap }_{\gamma \in \Gamma }{C}_{\gamma } \) . For each \( \gamma \in \Gamma \), we have \( x, y \in {C}_{\gamma } \), and since \( {C}_{\gamma } \) is convex, \( \left\lbrack {x, y}\right\rbrack \subseteq {C}... | Yes |
Theorem 4.11. Let \( A \neq \varnothing \) be a subset of an affine space \( E \) . Then \( \operatorname{co}\left( A\right) \) is a convex set; in fact, \( \operatorname{co}\left( A\right) \) is the smallest convex set containing \( A \) . | Proof. The proof is essentially a repeat of the proof of Lemma 4.2, but we now make the additional requirements that \( 0 < \alpha < 1 \) and that \( {\left\{ {\lambda }_{i}\right\} }_{1}^{k},{\left\{ {\mu }_{j}\right\} }_{1}^{l} \) be nonnegative in that proof. It suffices to note that all the affine combinations now ... | No |
Corollary 4.12. If \( C \) is a convex set in an affine space \( E \), then \( \operatorname{co}\left( C\right) = C \) , that is, all convex combinations of elements from \( C \) lie in \( C \), | \[ {\lambda }_{i} \geq 0,{x}_{i} \in C, i = 1,\ldots, k,\mathop{\sum }\limits_{{i = 1}}^{k}{\lambda }_{i} = 1\; \Rightarrow \;\mathop{\sum }\limits_{{i = 1}}^{k}{\lambda }_{i}{x}_{i} \in C. \] | Yes |
Theorem 4.13. (Carathéodory) Let \( A \) be a nonempty subset of an affine space \( E \) . Every element of \( \operatorname{co}\left( A\right) \) can be represented as a convex combination of affinely independent elements from \( A \) .\n\nConsequently, if \( n = \dim \left( {\operatorname{aff}\left( A\right) }\right)... | Proof. Let\n\n\[ x = \mathop{\sum }\limits_{{i = 1}}^{k}{\lambda }_{i}{x}_{i} \in \operatorname{co}\left( A\right) ,\text{ where }\mathop{\sum }\limits_{{i = 1}}^{k}{\lambda }_{i} = 1,\;{\lambda }_{i} > 0. \]\n\n(4.3)\n\nIf \( {\left\{ {x}_{i}\right\} }_{1}^{k} \) is affinely independent, then \( {\left\{ {x}_{i} - {x}... | Yes |
Corollary 4.15. If \( C \) is a nonempty compact subset of a finite-dimensional affine space \( E \), then so is the set \( \operatorname{co}\left( C\right) \) . | Proof. It follows from Theorem 4.13 that\n\n\[ \operatorname{co}\left( C\right) = \left\{ {\mathop{\sum }\limits_{{i = 1}}^{{n + 1}}{\lambda }_{i}{x}_{i} : {\lambda }_{i} \geq 0,{x}_{i} \in C, i = 1,\ldots, n + 1,\mathop{\sum }\limits_{{i = 1}}^{{n + 1}}{\lambda }_{i} = 1}\right\} ,\] \n\nwhere \( n = \dim \left( C\rig... | Yes |
Let \( E \) be an affine space in a normed vector space. If \( C \subseteq E \) is a convex set, then its closure \( \bar{C} \) is also a convex set. | To prove the first statement, define the convex set\n\n\[ \n{C}_{\epsilon } \mathrel{\text{:=}} \{ z : \parallel z - x\parallel < \epsilon, x \in C\} = \{ x + u : x \in C,\parallel u\parallel < \epsilon \} = C + {B}_{\epsilon }\left( 0\right) .\n\]\n\nNote that \( \bar{C} \mathrel{\text{:=}} { \cap }_{\epsilon > 0}{C}_... | No |
Lemma 4.18. A set \( K \) in a vector space \( E \) is a convex cone if and only if\n\n\[ x, y \in K\\text{ and }t > 0\; \\Rightarrow \;{tx} \in K, x + y \in K. \] | Proof. Let \( x, y \in K \) . If \( K \) is a convex cone, then \( \\left( {x + y}\\right) /2 \in K \), since \( K \) is convex, and \( x + y = 2\\left( {\\left( {x + y}\\right) /2}\\right) \in K \), since \( K \) is a cone. This proves (4.6). Conversely, if \( 0 < t < 1 \) and (4.6) holds, then \( \\left( {1 - t}\\rig... | No |
Theorem 4.20. Let \( A \) be a nonempty set in a vector space \( E \) . Then \( \operatorname{cone}\left( A\right) \) is the smallest convex cone containing \( A \) . If \( K \) is a convex cone, then \( \operatorname{cone}\left( K\right) = K \) . | This is proved in exactly the same way as Theorem 4.11. In fact, the proof here is somewhat simpler, since the weights \( \left\{ {\lambda }_{i}\right\} \) in a positive combination are not required to sum to one. | No |
Theorem 4.21. (Carathéodory) Let \( A \) be a nonempty subset of a vector space \( E \) . Every element of \( \operatorname{cone}\left( A\right) \) can be represented as a positive combination of linearly independent elements from \( A \) . Consequently, if \( n = \) \( \dim \left( {\operatorname{span}\left( A\right) }... | Proof. The proof is essentially the same as in the affine case. Let\n\n\[ x = \mathop{\sum }\limits_{{i = 1}}^{k}{\lambda }_{i}{x}_{i} \in \operatorname{cone}\left( A\right) ,\text{ where all }{\lambda }_{i} > 0. \]\n\n(4.7)\n\nIf the vectors \( {\left\{ {x}_{i}\right\} }_{1}^{k} \) are linearly dependent, then there e... | No |
Lemma 4.24. Let \( f : E \rightarrow \mathbb{R} \cup \{ \infty \} \) be a function on a vector space \( E \) . The function \( f \) is convex if and only if \( \operatorname{epi}\left( f\right) \subseteq E \times \mathbb{R} \) is a convex set in \( E \times \mathbb{R} \) . | Proof. First, assume that \( f \) is convex. Let \( \left( {{x}_{i},{\alpha }_{i}}\right) \in \operatorname{epi}\left( f\right), i = 1,2 \), and \( 0 \leq \lambda \leq 1 \) . We have \( f\left( {x}_{i}\right) \leq {\alpha }_{i}\left( {i = 1,2}\right) \), and since \( f \) is a convex function, we obtain\n\n\[ f\left( {... | Yes |
Corollary 4.25. (Jensen’s inequality) If \( f : E \rightarrow \mathbb{R} \cup \{ \infty \} \) is a convex function, then\n\n\[ f\left( {\mathop{\sum }\limits_{1}^{k}{\lambda }_{i}{x}_{i}}\right) \leq \mathop{\sum }\limits_{1}^{k}{\lambda }_{i}f\left( {x}_{i}\right) \]\n\nwhenever \( {\lambda }_{i} \geq 0, i = 1,\ldots,... | Proof. Since the corollary clearly holds if \( f\left( {x}_{i}\right) = \infty \) for some \( {x}_{i} \), we assume \( f\left( {x}_{i}\right) \in \mathbb{R} \) for all \( i = 1,\ldots, k \) . The points \( \left( {{x}_{i}, f\left( {x}_{i}\right) }\right) \) lie in \( \operatorname{epi}\left( f\right) \), and since \( \... | Yes |
Lemma 4.26. The following functions are convex:\n\n(a) The affine function \( f\left( x\right) = \langle a, x\rangle + b \), where \( a \in E \), and \( b \in \mathbb{R} \) . | Proof. The proofs of (a) and (c) are trivial. | No |
Theorem 4.27. Let \( C \) be a convex set in \( {\mathbb{R}}^{n} \), and let \( f \) be a Gâteaux differentiable function on an open set containing \( C \). Then \( f \) is convex on \( C \) if and only if the tangent plane at any point \( x \in C \) lies below the graph of \( f \), that is,\n\n\[ f\left( y\right) \geq... | Proof. First, assume that \( f \) is a convex function. If \( t \in \left( {0,1}\right) \), then the inequality \( f\left( {x + t\left( {y - x}\right) }\right) = f\left( {\left( {1 - t}\right) x + {ty}}\right) \leq \left( {1 - t}\right) f\left( x\right) + {tf}\left( y\right) \) can be\n\nwritten as\n\[ \frac{f\left( {x... | Yes |
Theorem 4.28. Let \( C \) be a convex set in \( {\mathbb{R}}^{n} \), and let \( f \) be a twice Fréchet differentiable function on an open set containing \( C \). Then,\n\n(a) The function \( f \) is convex on \( C \) if and only if the Hessian \( {Hf}\left( x\right) \) is positive semidefinite at every point \( x \in ... | Proof. (a). First, assume that \( f \) is convex and consider an arbitrary direction \( d \in E \). It follows from Theorem 4.27 and Taylor’s formula that\n\n\[ f\left( x\right) + t\langle \nabla f\left( x\right), d\rangle \leq f\left( {x + {td}}\right) \]\n\n\[ = f\left( x\right) + t\langle \nabla f\left( x\right), d\... | Yes |
Consider the quadratic function\n\n\[ f\left( x\right) = \frac{1}{2}\langle {Qx}, x\rangle + \langle c, x\rangle \]\n\nwhere \( Q \) is a symmetric \( n \times n \) matrix and \( c \in {\mathbb{R}}^{n} \). Then:\n\n(a) The function \( f \) is convex if and only if \( Q \) is positive semidefinite;\n\n(b) The function \... | Proof. We have \( \nabla f\left( x\right) = {Qx} + c \) and \( {Hf}\left( x\right) = Q \) ; see Example 1.26. Thus, Theorem 4.28 proves (a).\n\nIf \( f\left( x\right) \) is strictly convex and \( h \neq 0 \), then Theorem 4.27 gives\n\n\[ f\left( x\right) + \langle \nabla f\left( x\right), h\rangle < f\left( {x + h}\ri... | Yes |
Theorem 4.31. Let \( C \) be a convex set in \( {\mathbb{R}}^{n} \) such that \( \operatorname{int}\left( C\right) \neq \varnothing \), and let \( f : C \rightarrow \mathbb{R} \) be a convex function. If \( x \in \operatorname{int}\left( C\right) \), and the partial derivatives \( {\left\{ \partial f\left( x\right) /\p... | Proof. Define the function\n\n\[ g\left( h\right) \mathrel{\text{:=}} f\left( {x + h}\right) - f\left( x\right) - \langle \nabla f\left( x\right), h\rangle . \]\n\nNote that \( g \) is a convex function, \( g\left( 0\right) = 0 \), and \( \nabla g\left( 0\right) = 0 \) . We have\n\n\[ g\left( h\right) = g\left( {\frac{... | Yes |
Theorem 4.32. Let \( f : C \rightarrow \mathbb{R} \) be a convex function on a convex set \( C \) in a vector space \( E \) . Any local minimizer of \( f \) on \( C \) is a global minimizer of \( f \) on C. If \( f \) is strictly convex, then there exists at most one global minimizer of \( f \) on \( C \) . | Proof. Let \( {x}^{ * } \in C \) be a local minimizer of \( f \) on \( C \) . If \( x \in C \), then the line segment \( \left\lbrack {{x}^{ * }, x}\right\rbrack \) lies in \( C \) . For \( t \in \left( {0,1}\right) \), the point \( {x}_{t} \mathrel{\text{:=}} {x}^{ * } + t\left( {x - {x}^{ * }}\right) = \) \( \left( {... | Yes |
Theorem 4.33. Let \( C \) be a convex set in \( {\mathbb{R}}^{n} \), and let \( f \) be a Gâteaux differentiable function on an open set containing \( C \). (a) (First-order necessary condition for a local minimizer) If \( {x}^{ * } \in C \) is a local minimizer of \( f \) on \( C \), then \[ \left\langle {\nabla f\lef... | Proof. To prove (a), pick a point \( x \in C \) . Since \( C \) is convex, \( \left\lbrack {{x}^{ * }, x}\right\rbrack \subseteq C \), and since \( {x}^{ * } \) is a local minimizer of \( f \) on \( C \), we have \( f\left( {{x}^{ * } + t\left( {x - {x}^{ * }}\right) }\right) \geq f\left( {x}^{ * }\right) \) when \( t ... | Yes |
Let \( f \) be a Gâteaux differentiable function in a neighborhood of a convex set \( C \subseteq {\mathbb{R}}^{n} \). If \( C \) has nonempty interior, and \( {x}^{ * } \in \operatorname{int}\left( C\right) \) is a local minimizer of \( f \), then \( \nabla f\left( {x}^{ * }\right) = 0 \). | This equation also follows from the variational inequality, since choosing \( x = {x}^{ * } - \epsilon \nabla f\left( {x}^{ * }\right) \in C \) in (4.16) gives \( \begin{Vmatrix}{\nabla f\left( {x}^{ * }\right) }\end{Vmatrix} \leq 0 \). | No |
Consider the minimization of a differentiable function on an affine subspace,\n\n\\[ \n\\min \\;f\\left( x\\right) \n\\]\n\n\\[ \n\\text{s. t.}{Ax} = b\\text{,}\n\\]\n\nwhere \\( f : {\\mathbb{R}}^{n} \\rightarrow \\mathbb{R}, A \\) is an \\( m \\times n \\) matrix, and \\( b \\in {\\mathbb{R}}^{m} \\) . Define \\( C =... | Since \\( \\left\\{ {z = x - {x}^{ * } : x \\in C}\\right\\} = \\{ z : {Az} = 0\\} = N\\left( A\\right) \\), the variational inequality\n\nbecomes\n\n\\[ \n\\left\\langle {\\nabla f\\left( {x}^{ * }\\right), z}\\right\\rangle \\geq 0\\text{ for all }z \\in N\\left( A\\right) .\n\\]\n\nIf \\( z \\in N\\left( A\\right) \... | Yes |
Consider the quadratic program\n\n\\[ \min f\\left( x\\right) \\mathrel{\\text{:=}} \\frac{1}{2}\\langle {Qx}, x\\rangle + {c}^{T}x \\]\n\n\\[ \\text{s. t.}x \\geq 0\\text{,}\\]\n\nwhere \\( Q \\) is an \\( n \\times n \\) symmetric and \\( c \\in {\\mathbb{R}}^{n} \\) .\n\nIf \\( Q \\) is positive definite, then the o... | Let \\( {x}^{ * } \\geq 0 \\) be a local minimizer of \\( f \\) on the nonnegative orthant. Since \\( \\nabla f\\left( {x}^{ * }\\right) = Q{x}^{ * } + c \\), the variational inequality becomes\n\n\\[ \\left\\langle {Q{x}^{ * } + c, x - {x}^{ * }}\\right\\rangle \\geq 0\\text{ for all }x \\geq 0\\text{ in }{\\mathbb{R}... | Yes |
Consider the maximization problem\n\n\\[ \n\\max \\;g\\left( x\\right) \\mathrel{\\text{:=}} {x}_{1}^{{\\alpha }_{1}}\\ldots {x}_{n}^{{\\alpha }_{n}}, \n\\]\n\n\\[ \n\\text{s. t.}{x}_{1} + \\cdots + {x}_{n} = 1 \n\\]\n\n\\[ \n{x}_{i} \\geq 0,\\;i = 1,\\ldots, n \n\\] | Since each \\( {x}_{i}^{ * } \\) must clearly be positive at a local maximizer, we can reformulate the problem:\n\n\\[ \n\\min f\\left( x\\right) \\mathrel{\\text{:=}} - {\\alpha }_{1}\\ln {x}_{1} + \\cdots + \\left( {-{\\alpha }_{n}}\\right) \\ln {x}_{n} \n\\]\n\n\\[ \n\\text{s. t.}{x}_{1} + \\cdots + {x}_{n} = 1.\\te... | Yes |
Finally, we consider the minimization of a differentiable function on a convex polyhedron,\n\n\\[ \n\\min \\;f\\left( x\\right) \n\\]\n\n\\[ \n\\text{s. t.}{Ax} \\leq a\\text{,}\n\\]\n\n\\[ \n{Bx} = b, \n\\]\n\nwhere \\( f : {\\mathbb{R}}^{n} \\rightarrow \\mathbb{R}, A \\) and \\( B \\) are \\( m \\times n \\) and \\(... | It is not a trivial matter to rewrite this system of potentially infinitely many conditions (one condition for each \\( x \\in P \\) ) in a compact, manageable form, but it is possible. Note that the implication above is equivalent to stating that the linear inequality system\n\n\\[ \n{Ax} \\leq a,\\;{Bx} = b,\\;\\left... | Yes |
Theorem 4.41. Let \( C \subseteq {\mathbb{R}}^{n} \) be a closed convex set, and \( f \) a lower semicontinuous, Gâteaux differentiable function in a neighborhood of \( C \). If \( f \) is bounded from below on \( C \), then there exists a sequence \( {\left\{ {x}_{k}\right\} }_{k = 1}^{\infty } \) of points in \( C \)... | Proof. Applying Theorem 3.2 with \( \lambda = 1 \) and \( \epsilon = 1/k \), we obtain a point \( {x}_{k} \in C \) satisfying the conditions \[ f\left( {x}_{k}\right) \leq \mathop{\inf }\limits_{C}f + \frac{1}{k}\;\text{ and }\;f\left( {x}_{k}\right) \leq f\left( x\right) + \frac{\begin{Vmatrix}x - {x}_{k}\end{Vmatrix}... | Yes |
Lemma 5.2. If \( C \) is a convex set in an affine space \( A \), then \( \operatorname{ai}\left( C\right) \) and \( \operatorname{ac}\left( C\right) \) are also convex sets in \( A \) . | Proof. Let \( x, y \in \operatorname{ai}\left( C\right) \) . If \( u \in A \), then there exists \( \delta > 0 \) such that \( x + \delta (u - \) \( x)\rbrack = : \left\lbrack {x, p}\right\rbrack \subset C \) and \( \left\lbrack {y, y + \delta \left( {u - y}\right) }\right\rbrack = : \left\lbrack {y, q}\right\rbrack \s... | Yes |
Lemma 5.3. If \( C \) is a convex set in a finite-dimensional affine space \( A \), then \( \operatorname{rai}\left( C\right) \neq \varnothing \) . | Proof. Let \( {\left\{ {x}_{i}\right\} }_{1}^{k} \subseteq C \) be an affine basis of \( \operatorname{aff}\left( C\right) \), and let \( x \mathrel{\text{:=}} \left( {\mathop{\sum }\limits_{1}^{k}{x}_{i}}\right) /k \) be the center of the simplex defined by \( {\left\{ {x}_{i}\right\} }_{1}^{k} \) . We claim that \( x... | Yes |
Lemma 5.5. Let \( C \) be a convex set in an affine space \( A \) . If \( y \in \operatorname{ac}\left( C\right) \) and \( x \in \operatorname{ai}\left( C\right) \), then \( \lbrack x, y) \subset \operatorname{ai}\left( C\right) \) . | Proof. First, assume that \( y \in C \) . Let \( z \mathrel{\text{:=}} {tx} + \left( {1 - t}\right) y = y + t\left( {x - y}\right) \in C \) , \( t \in \left( {0,1}\right) \) ; see the last figure in Figure 5.1. We claim that \( z \in \operatorname{ai}\left( C\right) \) . Let \( d \mathrel{\text{:=}} u - x \) be an arbi... | Yes |
Theorem 5.7. Let \( C \) be a convex set in a vector space \( E \) such that \( 0 \in \operatorname{rai}\left( C\right) \) . The gauge function \( {p}_{C} \) is a nonnegative extended-valued function, \( {p}_{C} : E \rightarrow \) \( \mathbb{R} \cup \{ + \infty \} \), that is finite-valued precisely on the linear subsp... | Proof. Evidently, \( p\left( x\right) \mathrel{\text{:=}} {p}_{C}\left( x\right) \geq 0 \) for all \( x \in E \) . Since \( 0 \in \operatorname{rai}\left( C\right) \), there exists \( t > 0 \) such that \( x \in {tC} \) if and only if \( x \in L \mathrel{\text{:=}} \operatorname{span}\left( C\right) \) . Thus \( p\left... | Yes |
Corollary 5.9. Let \( C \) be a convex algebraic body in an affine space \( A \) . Then\n\n\[ \operatorname{rai}\left( C\right) = \operatorname{rai}\left( {\operatorname{rai}\left( C\right) }\right) = \operatorname{rai}\left( {\operatorname{ac}\left( C\right) }\right) = \{ x \in A : p\left( x\right) < 1\} \]\n\n\[ \ope... | Proof. Let \( {x}_{0} \in \operatorname{rai}\left( C\right) \) . Evidently, \( 0 \in \operatorname{rai}\left( {C - {x}_{0}}\right) = \operatorname{rai}\left( C\right) - {x}_{0},\operatorname{ac}\left( {C - {x}_{0}}\right) = \) \( \operatorname{ac}\left( C\right) - {x}_{0} \), and \( {p}_{C}\left( x\right) = {p}_{\left(... | Yes |
Lemma 5.10. Let \( {\left\{ {C}_{i}\right\} }_{1}^{m} \) be convex sets in a vector space \( E \) . If \( { \cap }_{1}^{m}\operatorname{rai}\left( {C}_{i}\right) \neq \) \( \varnothing \) , then\n\n(a) \( \operatorname{aff}\left( {{ \cap }_{1}^{m}{C}_{i}}\right) = { \cap }_{1}^{m}\operatorname{aff}\left( {C}_{i}\right)... | Proof. Write \( C \mathrel{\text{:=}} { \cap }_{1}^{m}{C}_{i} \) and \( D \mathrel{\text{:=}} { \cap }_{1}^{m}\operatorname{aff}\left( {C}_{i}\right) \) . Since \( C \subseteq D \) and \( D \) is an affine set, it follows that \( \operatorname{aff}\left( C\right) \subseteq D \) . To prove the reverse inclusion, let \( ... | Yes |
Lemma 5.11. Let \( C \) and \( D \) be two convex sets in a vector space \( E \) . If \( \operatorname{rai}\left( C\right) \neq \) \( \varnothing \) and \( \operatorname{rai}\left( D\right) \neq \varnothing \), then \( \operatorname{rai}\left( {C + D}\right) = \operatorname{rai}\left( C\right) + \operatorname{rai}\left... | Proof. Let \( x \in \operatorname{rai}\left( C\right) \) and \( y \in \operatorname{rai}\left( D\right) \) . Given arbitrary points \( u \in C \) and \( v \in D \), there exist \( {u}_{1} \in C \) and \( {v}_{1} \in D \) such that \( x \in \left( {u,{u}_{1}}\right) \) and \( y \in \left( {v,{v}_{1}}\right) \) . We may ... | Yes |
Corollary 5.14. Let \( E \) and \( F \) be affine spaces, and \( A : E \rightarrow F \) a multivalued map whose graph \( \operatorname{gr}\left( A\right) = C \) is a nonempty convex set in \( E \times F \) . Then\n\n\[ \operatorname{dom}\operatorname{rai}\left( {\operatorname{gr}\left( A\right) }\right) \subseteq \oper... | Proof. The inclusion follows immediately from Lemma 5.13. If \( x \in \operatorname{rai}\operatorname{dom}\left( A\right) \) and \( y \in \operatorname{rai}A\left( x\right) \neq \varnothing \), then Lemma 5.13 implies that \( \left( {x, y}\right) \in \operatorname{raigr}\left( A\right) \), and we have \( x \in \operato... | Yes |
Corollary 5.15. Let \( {C}_{i} \) be a convex set in an affine space \( {A}_{i}, i = 1,\ldots, k \) . If each \( \operatorname{rai}\left( {C}_{i}\right) \) is nonempty, then\n\n\[ \operatorname{rai}\left( {{C}_{1} \times {C}_{2} \times \cdots \times {C}_{k}}\right) = \operatorname{rai}\left( {C}_{1}\right) \times \oper... | Proof. The proof is trivial for \( k = 1 \) and follows immediately from Lemma 5.13 for \( k = 2 \) . The proof is easily completed by induction on \( k \) . | No |
Lemma 5.16. Let \( A : E \rightarrow F \) be a multivalued affine map between two affine spaces \( E \) and \( F \), and \( C \subseteq E \) a convex set such that \( \operatorname{rai}\left( C\right) \cap \operatorname{dom}\left( A\right) \neq \varnothing \) . Then we always have \[ A\left( {\operatorname{rai}\left( C... | Proof. Consider the multivalued map \( B : F \rightarrow E \) whose graph is the convex set \[ \operatorname{gr}\left( B\right) \mathrel{\text{:=}} \operatorname{gr}\left( A\right) \cap \left( {C \times F}\right) \] and note that \[ \operatorname{dom}\left( B\right) = \{ y : \exists x \in C, y \in A\left( x\right) \} =... | Yes |
Lemma 5.18. Let \( C \) be a convex set in a topological affine space \( A \) with a nonempty interior. If \( x \in \operatorname{int}\left( C\right) \) and \( y \in \bar{C} \), then \( \lbrack x, y) \subseteq \operatorname{int}\left( C\right) \) . Consequently, \( \operatorname{int}\left( C\right) \) is a convex set. | Proof. Let \( z \mathrel{\text{:=}} y + t\left( {x - y}\right), t \in \left( {0,1}\right) \) . We claim that \( z \in \operatorname{int}\left( C\right) \) . Let \( U \subset C \) be a neighborhood of \( x \) ; see Figure 5.3. Since \( y = \left( {z - {tx}}\right) /\left( {1 - t}\right) \in \) \( \left( {z - {tU}}\right... | Yes |
Lemma 5.19. Let \( C \) be a nonempty convex set in a topological affine space A. Then \( \bar{C} \) is a convex set, and \( \operatorname{ac}\left( C\right) \subseteq \bar{C} \) . | Proof. Let \( x, y \in \bar{C} \) and \( z \mathrel{\text{:=}} y + t\left( {x - y}\right) = {tx} + \left( {1 - t}\right) y, t \in \left( {0,1}\right) \) . We claim that \( z \in \bar{C} \) . Let \( {U}_{z} \) be a neighborhood of \( z \) . Since the map \( \left( {u, v}\right) \mapsto \) \( {tu} + \left( {1 - t}\right)... | Yes |
Theorem 5.20. If \( C \) is a convex body in a topological affine space \( A \), that is, \( \operatorname{int}\left( C\right) \neq \varnothing \), then\n\n\[ \operatorname{int}\left( C\right) = \operatorname{int}\left( \bar{C}\right) = \operatorname{ai}\left( C\right) \;\text{ and }\;\bar{C} = \overline{\operatorname{... | Proof. To prove the equality \( \operatorname{int}\left( C\right) = \operatorname{ai}\left( C\right) \), it suffices to prove the claim that \( \operatorname{ai}\left( C\right) \subseteq \operatorname{int}\left( C\right) \), since the reverse inclusion is already proved in Lemma 5.18. Let \( x \in \operatorname{ai}\lef... | Yes |
Lemma 5.22. If \( C \) is a nonempty convex set in a finite-dimensional affine space \( A \), then \( \operatorname{ri}\left( C\right) \neq \varnothing \) . Moreover, \( \dim \left( C\right) = \dim \left( {\operatorname{ri}\left( C\right) }\right) \) . | Proof. Let \( {\left\{ {x}_{i}\right\} }_{1}^{k} \subset C \) be an affine basis for aff \( \left( C\right) \) . Consider the simplices\n\n\[ S \mathrel{\text{:=}} \left\{ {\mathop{\sum }\limits_{{i = 1}}^{k}{t}_{i}{x}_{i} : \mathop{\sum }\limits_{1}^{k}{t}_{i} = 1,{t}_{i} > 0, i = 1,\ldots, k}\right\} \subset C,\]\n\n... | Yes |
Theorem 5.23. If \( C \) is a nonempty finite-dimensional convex set, then \( \operatorname{ri}\left( C\right) \neq \varnothing \), and | \[ \operatorname{ri}\left( C\right) = \operatorname{ri}\left( \bar{C}\right) = \operatorname{rai}\left( C\right) \;\text{ and }\;\bar{C} = \overline{\operatorname{ri}\left( C\right) } = \operatorname{ac}\left( C\right) .\ | No |
Lemma 5.25. Let \( F \) and \( C \) be two convex sets such that \( F \subset C \) . Then \( F \) is a face of \( C \) if and only if \( C \smallsetminus F \) is a convex set. | This is an easy consequence of the definition of a face, and can be used as an alternative definition of a face. | No |
Lemma 5.27. Let \( C \) be a convex set in a vector space \( E, F \) a face of \( C \), and \( D \) a convex subset of \( C \) . If \( \operatorname{ri}\left( D\right) \cap F \neq \varnothing \), then \( D \subseteq F \) . | Proof. Pick \( z \in \operatorname{ri}\left( D\right) \cap F \) . If \( x \in D \), then \( z \in \operatorname{ri}\left( D\right) \) implies that there exists \( y \in D \) such that \( z \in \left( {x, y}\right) \) . Since \( F \) is a face of \( C \), we have \( x \in F \) . | Yes |
Lemma 5.28. Let \( C \) be a closed convex set in a vector space \( E \). If \( 0 \neq d \in E \) is a recession direction of \( C \), then \( q + {\mathbb{R}}_{ + }d \subseteq C \) for every \( q \in C \). | Proof. Suppose that \( p + {\mathbb{R}}_{ + }d \subseteq C \), and let \( q \in C, q \neq p \). It is easy to see geometrically that the convex hull of \( q \) and \( p + R \) is the union of sets \( \{ q\} \) and \( \lbrack p, q) + R \), whose closure is the set \( \left\lbrack {p, q}\right\rbrack + R \) ; see Figure ... | No |
Theorem 5.30. Let \( C \) be a nonempty, closed, convex set in a finite-dimensional linear space \( E \). Then \( C \) contains an affine face \( F \). | Proof. We first demonstrate the existence of an affine face by induction on the dimension of \( C \). Suppose \( \dim \left( C\right) = n \) and that we have proved the existence of affine faces for convex sets with dimension less than \( n \). Clearly, we may assume that \( C \) is full-dimensional, that is, \( E \) h... | Yes |
Lemma 5.33. Let \( C \) be a nonempty, closed, convex set in a finite-dimensional linear space \( E \), and let \( M \) be a linear subspace of \( E \) complementary to \( {L}_{C} \). The set \( C \) can decomposed as \[ C = \widehat{C} \oplus {L}_{C} \] where \( \widehat{C} \) is a line-free, closed convex set in \( M... | Proof. Define \( \widehat{C} \mathrel{\text{:=}} C \cap M \) . If \( x \in C \), we can write \( x = l + m \), where \( l \in {L}_{C} \) and \( m \in M \) ; then we have \( m \in \widehat{C} \), since \( m \in M \), and \( m = x - l \in C + {L}_{C} = C \) by virtue of Theorem 5.30. This shows that \( \widehat{C} \neq \... | Yes |
Lemma 5.34. Let \( C \) be a closed convex set \( C \) in a finite-dimensional linear space \( E \). The relative boundary \( \operatorname{rbd}\left( C\right) \mathrel{\text{:=}} C \smallsetminus \operatorname{ri}\left( C\right) \) of \( C \) is convex if and only if \( C \) is either an affine subspace of \( E \) or ... | Proof. If \( C \) is either an affine subset or the intersection of an affine subspace with a closed half-space, then it is clear that \( \operatorname{rbd}\left( C\right) \) is convex.\n\nTo prove the converse, assume without loss of generality that \( C \) is full-dimensional. If the boundary of \( C \) is empty, the... | Yes |
Lemma 5.35. Let \( C \) be a nonempty, closed, convex set in a finite-dimensional linear space \( E \) . If \( C \) is not an affine subspace or the intersection of an affine subspace with a closed half-space, then every point in the relative interior of \( C \) lies on a line segment whose endpoints lie on the relativ... | Proof. Again, we may assume that \( C \) is full-dimensional. Since \( \partial C \) is not convex by Lemma 5.34, there exist two points \( x, y \in \partial C \) such that \( \left\lbrack {x, y}\right\rbrack \) intersects \( \operatorname{int}\left( C\right) \) . It follows from Lemma 5.28 that the line passing throug... | Yes |
Theorem 5.36. A nonempty, line-free, closed, convex set \( C \) in a finite-dimensional linear space \( E \) is the convex hull of its extreme points and extreme rays, that is, any point \( x \in C \) has a representation \[ x = \mathop{\sum }\limits_{{i = 1}}^{k}{\lambda }_{i}{v}_{i} + \mathop{\sum }\limits_{{j = 1}}^... | Proof. We use induction on the dimension of \( C \) . Suppose \( \dim \left( C\right) = n \) and that we have proved the theorem for convex sets with dimension less than \( n \) . First, suppose that \( x \in \operatorname{rbd}\left( C\right) = C \smallsetminus \operatorname{ri}\left( C\right) \) . By Theorem 6.8, ther... | Yes |
Theorem 5.37. Let \( C \) be a closed convex set \( C \) in a finite-dimensional vector space \( E \) . The set \( C \) can be decomposed as\n\n\[ C = {L}_{C} \oplus \widehat{C} \]\n\nwhere \( {L}_{C} \) is a linear subspace and \( \widehat{C} \) is a line-free convex set lying in a linear subspace complementary to \( ... | Proof. The theorem follows immediately from Lemma 5.33 and Theorem 5.36. | Yes |
Lemma 5.39. Let \( C = {C}_{1} + {C}_{2} \), where \( {C}_{1} \) and \( {C}_{2} \) are nonempty convex sets. If \( F \) is a face of \( C \), then there exist faces \( {F}_{i} \) of \( {C}_{i}, i = 1,2 \), such that \( F = {F}_{1} + {F}_{2} \) | Proof. Define the sets\n\n\[ \n{F}_{1} = \left\{ {x \in {C}_{1} : \exists y \in {C}_{2}, x + y \in F}\right\} ,\;{F}_{2} = \left\{ {x \in {C}_{2} : \exists x \in {C}_{1}, x + y \in F}\right\} . \n\]\n\nIt is easy to verify that \( {F}_{1} \) and \( {F}_{2} \) are convex sets. We claim that \( {F}_{1} \) is a face of \(... | Yes |
Lemma 5.40. Let \( C = {C}_{1} + {C}_{2} \), where \( {C}_{1} \) and \( {C}_{2} \) are nonempty convex sets. If \( z \) is an extreme point of \( C \), then \( z \) has a unique representation \( z = x + y \) , where \( x \in {C}_{1}, y \in {C}_{2} \) . Moreover, in this representation \( x \) is an extreme point of \(... | Proof. It follows from Lemma 5.39 that \( z = x + y \), where \( x \) and \( y \) are extreme points of \( {C}_{1} \) and \( {C}_{2} \), respectively. If \( z = \bar{x} + \bar{y} \), where \( \bar{x} \in {C}_{1},\bar{y} \in {C}_{2} \), then we have \( z = \left( {x + \bar{y}}\right) /2 + \left( {\bar{x} + y}\right) /2 ... | Yes |
Lemma 5.41. If \( C \neq \varnothing \) is a closed convex set in a finite-dimensional vector space \( E \), then \[ \overline{K\left( C\right) } = K\left( C\right) \cup \{ \left( {d,0}\right) : d \in \operatorname{rec}\left( C\right) \} . \] | Proof. Denote the set on the right-hand side by \( D \) . Let \( d \in \operatorname{rec}\left( C\right) \) . If \( x \in C \) , then \( {x}_{n} \mathrel{\text{:=}} x + {nd} \in C,\left( {{x}_{n},1}\right) /n \in K\left( C\right) \), and \( \left( {{x}_{n},1}\right) /n \rightarrow \left( {d,0}\right) \in \overline{K\le... | Yes |
Lemma 5.42. Let \( f : E \rightarrow \mathbb{R} \cup \{ \infty \} \) be a convex function in a normed linear space \( E \) such that \( \operatorname{dom}\left( f\right) \) has a nonempty interior. If \( f \) is bounded from above in a neighborhood of some point \( {x}_{0} \in \operatorname{int}\left( {\operatorname{do... | Proof. Let \( {B}_{{r}_{0}}\left( {x}_{0}\right) \in \operatorname{int}\left( {\operatorname{dom}\left( f\right) }\right) \) be such that \( f\left( x\right) \leq a < \infty \) for \( x \in {B}_{{r}_{0}}\left( {x}_{0}\right) \) , and let \( x \in \operatorname{int}\left( {\operatorname{dom}\left( f\right) }\right) \) b... | Yes |
Theorem 5.43. Let \( f : E \rightarrow \mathbb{R} \cup \{ \infty \} \) be a convex function on a normed linear space \( E \) . Then \( f \) is continuous in \( \operatorname{ri}\left( {\operatorname{dom}\left( f\right) }\right) \) if and only if there exists a point \( {x}_{0} \in \operatorname{ri}\left( {\operatorname... | Proof. Define \( C = \operatorname{dom}\left( f\right) \) and assume that \( \operatorname{int}\left( {\operatorname{dom}\left( f\right) }\right) \neq \varnothing \), by restricting \( f \) to \( \operatorname{aff}\left( {\operatorname{dom}\left( f\right) }\right) \) if necessary. If \( f \) is continuous at \( {x}_{0}... | Yes |
Theorem 5.45. Let \( f : E \rightarrow \mathbb{R} \cup \{ \infty \} \) be a convex function on a normed linear space \( E \) . If there exists a point \( {x}_{0} \in \operatorname{ri}\left( {\operatorname{dom}\left( f\right) }\right) \) such that \( f \) is bounded from above in a relative neighborhood of \( {x}_{0} \)... | Proof. We may again assume that \( \operatorname{int}\left( {\operatorname{dom}\left( f\right) }\right) \neq \varnothing \) . By virtue of Theorem \( {5.43}, f \) is continuous in a neighborhood \( {\bar{B}}_{{r}_{0}}\left( {x}_{0}\right) \) . Suppose that\n\n\[ m \leq f\left( x\right) \leq M\;\text{ on }\;{\bar{B}}_{{... | Yes |
Lemma 5.46. Let \( f : E \rightarrow \mathbb{R} \cup \{ \infty \} \) be a convex function on a finite-dimensional normed linear space \( E \) . If \( x \in \operatorname{ri}\left( {\operatorname{dom}\left( f\right) }\right) \), then \( f \) is bounded from above in a relative neighborhood of \( x \) . | Proof. Let \( n = \dim \left( E\right) \) . As in the proof of Theorem 5.43, we may assume that \( \operatorname{int}\left( {\operatorname{dom}\left( f\right) }\right) \neq \varnothing \) . If \( x \in \operatorname{int}\left( {\operatorname{dom}\left( f\right) }\right) \), then\n\n\[ x \in N = \operatorname{int}\left(... | Yes |
Theorem 6.1. Let \( C \subseteq {\mathbb{R}}^{n} \) be a nonempty closed convex set. The projection \( {\Pi }_{C}\left( x\right) \) of \( x \) onto \( C \) is characterized by the variational inequality\n\n\[ \left\langle {x - {\Pi }_{C}\left( x\right), z - {\Pi }_{C}\left( x\right) }\right\rangle \leq 0\text{ for all ... | Proof. Consider the minimization problem\n\n\[ \mathop{\min }\limits_{{z \in C}}f\left( z\right) \mathrel{\text{:=}} \frac{1}{2}\parallel z - x{\parallel }^{2} \]\n\nThe function \( f \) is clearly coercive on \( {\mathbb{R}}^{n} \), so that any sublevel set \( {l}_{\alpha }\left( f\right) \mathrel{\text{:=}} \) \( \{ ... | Yes |
Corollary 6.3. The function \( {\Pi }_{C} : {\mathbb{R}}^{n} \rightarrow C \) is nonexpansive, that is,\n\n\[ \n\begin{Vmatrix}{{\Pi }_{C}\left( {x}_{2}\right) - {\Pi }_{C}\left( {x}_{1}\right) }\end{Vmatrix} \leq \begin{Vmatrix}{{x}_{2} - {x}_{1}}\end{Vmatrix}\text{ for all }{x}_{1},{x}_{2} \in {\mathbb{R}}^{n}.\n\]\n... | Proof. The variational inequality (6.1) gives\n\n\[ \n\left\langle {{x}_{1} - {\Pi }_{C}\left( {x}_{1}\right) ,{\Pi }_{C}\left( {x}_{2}\right) - {\Pi }_{C}\left( {x}_{1}\right) }\right\rangle \leq 0,\n\]\n\n\[ \n\left\langle {{x}_{2} - {\Pi }_{C}\left( {x}_{2}\right) ,{\Pi }_{C}\left( {x}_{1}\right) - {\Pi }_{C}\left( ... | Yes |
Theorem 6.7. If \( C \subset {\mathbb{R}}^{n} \) is a nonempty convex set and \( \bar{x} \notin \operatorname{ri}\left( C\right) \), then there exists a hyperplane \( {H}_{\left( a,\alpha \right) } \) such that \( \bar{x} \in {H}_{\left( a,\alpha \right) } \) and \( C \subseteq {\bar{H}}_{\left( a,\alpha \right) }^{ + ... | Proof. We first assume that \( \bar{x} \notin \bar{C} \) . The variational inequality (6.1) gives\n\n\[ \left\langle {{\Pi }_{\bar{C}}\bar{x} - \bar{x}, x - {\Pi }_{\bar{C}}\bar{x}}\right\rangle \geq 0\text{ for all }x \in \bar{C}; \]\n\ndefining \( a \mathrel{\text{:=}} {\Pi }_{C}\left( \bar{x}\right) - \bar{x} \neq 0... | Yes |
Theorem 6.9. Let \( C \) and \( D \) be two nonempty convex sets in \( {\mathbb{R}}^{n} \) . If \( C \) and \( D \) are disjoint, then there exists a hyperplane \( {H}_{\left( a,\alpha \right) } \) that separates \( C \) and \( D \) , that is,\n\n\[ \langle a, x\rangle \leq \alpha \leq \langle a, y\rangle \text{ for al... | Proof. The trick is to define the set\n\n\[ A \mathrel{\text{:=}} C - D = \{ x - y : x \in C, y \in D\} ,\]\n\nand note that \( 0 \notin A \), due to \( C \cap D = \varnothing \) . The set \( A \) is convex, because \( A = \) \( C + \left( {-D}\right) \) is the Minkowski sum of the convex sets \( C \) and \( - D \) ; i... | Yes |
Theorem 6.10. (Strong separation theorem) Let \( C, D \) be two nonempty, disjoint, closed, convex sets in \( {\mathbb{R}}^{n} \). If one of them is compact, then \( C \) and \( D \) can be strongly separated. | Proof. Let us assume for definiteness that \( D \) is compact. We note that the theorem is equivalent to the existence of a hyperplane \( {H}_{\left( a,\alpha \right) } \) satisfying the condition\n\n\[ \mathop{\inf }\limits_{{x \in C}}\langle a, x\rangle > \alpha > \mathop{\max }\limits_{{y \in D}}\langle a, y\rangle ... | Yes |
Theorem 6.12. If \( C \subseteq {\mathbb{R}}^{n} \) is a nonempty closed convex set, then \( C \) is the intersection of all the closed half-spaces containing it, that is,\n\n\[ C = \mathop{\bigcap }\limits_{\left( a,\alpha \right) }\left\{ {{\bar{H}}_{\left( a,\alpha \right) }^{ + } : C \subseteq {\bar{H}}_{\left( a,\... | Proof. Denote by \( D \) the intersection set above. It is clear that \( D \) is a closed, convex set containing \( C \), so it remains to show that \( D \subseteq C \) .\n\nIf this is not true, then there exists a point \( {x}_{0} \in D \) that does not lie in \( C \) . Applying Theorem 6.10 to the convex sets \( \lef... | Yes |
Lemma 6.14. Two nonempty convex sets \( C \) and \( D \) in \( {\mathbb{R}}^{n} \) can be properly separated if and only if the origin and the convex set \( K \mathrel{\text{:=}} C - D \) can be properly separated. | Proof. Let \( H \mathrel{\text{:=}} {H}_{\left( a,\alpha \right) } \) be a hyperplane properly separating \( C \) and \( D \) such that \( C \subseteq {\bar{H}}^{ + }, D \subseteq {\bar{H}}^{ - } \), and assume without loss of generality that \( C \) does not lie on \( H \) . Then\n\n\[ \langle a, x\rangle \geq \alpha ... | Yes |
Theorem 6.15. (Proper separation theorem) Two nonempty convex sets \( C \) and \( D \) in \( {\mathbb{R}}^{n} \) can be properly separated if and only if \( \operatorname{ri}\left( C\right) \) and \( \operatorname{ri}\left( D\right) \) are disjoint. | Proof. Define the convex set \( K \mathrel{\text{:=}} C - D \) . It follows from Lemma 5.11 and Corollary 5.21 that \( \operatorname{ri}\left( K\right) = \operatorname{ri}\left( {C - D}\right) = \operatorname{ri}\left( C\right) - \operatorname{ri}\left( D\right) \) ; thus, \( \operatorname{ri}\left( C\right) \cap \oper... | Yes |
Theorem 6.16. Let \( C \) be a nonempty convex set in \( {\mathbb{R}}^{n} \). If \( D \) is a nonempty convex subset of the relative boundary of \( C\left( {D \subseteq \bar{C} \smallsetminus \operatorname{ri}\left( C\right) }\right) \), then there exists a support hyperplane to \( C \) containing \( D \) but not all p... | Proof. Since \( \operatorname{ri}\left( C\right) \cap \operatorname{ri}\left( D\right) \subseteq \operatorname{ri}\left( C\right) \cap D = \varnothing \), Theorem 6.15 implies that there exists a hyperplane \( H \mathrel{\text{:=}} {H}_{\left( a,\alpha \right) } \) properly separating \( C \) and \( D \), say\n\n\[ \la... | Yes |
Theorem 6.17. Let \( C \subset {\mathbb{R}}^{n} \) be a nonempty convex set. If \( M \) is an affine set such that \( \operatorname{ri}\left( C\right) \) and \( M \) are disjoint, then \( M \) can be extended to a hyperplane \( H \) such that \( \operatorname{ri}\left( C\right) \) and \( H \) are disjoint. | Proof. Since \( M \) is affine, \( \operatorname{ri}\left( M\right) = M \), and Theorem 6.15 implies that there exists a hyperplane \( H \) properly separating \( C \) and \( M \) . If \( M \subseteq H \), we are done; if not, there exists a point \( x \in M \smallsetminus H \neq \varnothing \) . The affine sets \( H \... | Yes |
Theorem 6.19. If \( K \subseteq {\mathbb{R}}^{n} \) is a closed convex cone, then \( K = {K}^{* * } \) . | Proof. From the definition of \( {K}^{ * } \), we see that if \( x \in K \), then \( \langle x, y\rangle \leq 0 \) for all \( y \in {K}^{ * } \) ; this proves that \( K \subseteq {K}^{* * } \) . Suppose that the reverse inclusion \( {K}^{* * } \subseteq \) \( K \) is not true, and pick a point \( x \in {K}^{* * } \smal... | Yes |
Theorem 6.20. Let \( C \subset {\mathbb{R}}^{n} \) be a nonempty open convex set. The distance function\n\n\[ \n{d}_{C}\left( x\right) \mathrel{\text{:=}} d\left( {x,\partial C}\right) = \min \{ \parallel x - z\parallel : z \in \partial C\}\n\]\n\nis a concave function on \( C \) that vanishes on the boundary of \( C \... | Proof. Let \( x \in C \) . If \( {H}^{ + } is an open half-space containing \( C \), then it is a simple consequence of the definition of \( {d}_{C} \) that \( {d}_{C}\left( x\right) \leq {d}_{{H}^{ + }}\left( x\right) \) ; thus\n\n\[ \n{d}_{C}\left( x\right) \leq \mathop{\inf }\limits_{{C \subseteq {H}^{ + }}}{d}_{{H}... | Yes |
Theorem 6.21. Let \( C \) and \( P \) be nonempty convex sets in \( {\mathbb{R}}^{n} \) such that \( P \) is a convex polyhedron, that is, \( P \) is the intersection of finitely many closed half-spaces in \( {\mathbb{R}}^{n} \). There exists a hyperplane separating \( C \) and \( P \) properly and not containing \( C ... | Proof. Suppose that there exists a hyperplane \( H \) that separates \( C \) and \( P \) and does not contain \( C \). If \( x \in \operatorname{ri}\left( C\right) \cap P \neq \varnothing \) as well, then we have \( x \in H \), and if \( y \in C \smallsetminus H \), then the line segment \( \left\lbrack {y, x}\right\rb... | Yes |
Lemma 6.22. Let \( {\left\{ {C}_{i}\right\} }_{1}^{k + 1}, k \geq 1 \), be convex sets in \( {\mathbb{R}}^{n} \) such that \( 0 \in \overline{{C}_{i}} \) for \( i = 1,\ldots, k + 1 \) . Consider the conditions\n\n(a) \( { \cap }_{i = 1}^{k + 1}{C}_{i} = \varnothing \) .\n\n(b) There exists \( l \mathrel{\text{:=}} \lef... | Proof. (a) implies (b): The important idea here is to define the sets\n\n\[ {K}_{1} \mathrel{\text{:=}} \left\{ {\left( {{x}_{k + 1},\ldots ,{x}_{k + 1}}\right) : {x}_{k + 1} \in {C}_{k + 1}}\right\} \]\n\n\[ {K}_{2} \mathrel{\text{:=}} {C}_{1} \times \cdots \times {C}_{k} \]\n\nand to note that \( {K}_{1} \cap {K}_{2}... | Yes |
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